a.math解方程α同β 10分!!

Q1. α+β= 2k- 1
αβ= k²
(a)
α-β= 開方(α^2 - 2αβ+β^2)
α-β= 開方(α^2 + 2αβ+β^2 - 4αβ)
α-β= 開方[(α+β)^2 - 4αβ]
α-β= 開方[(2k-2)^2 - 4k²]
α-β= 開方(4k² - 8k +4 - 4k²)
α-β= 開方(-8k + 4)
α<β
α-β<0
開方(-8k + 4)<0
-開方(8k - 4)<0
8k-4>0
k>1/2

(b) αβ= k²
α²β²= k^4

α-β= 開方(-8k + 4)

Q2.(a)α+β= 2k
α ．β=k-2

(b)(α+1)(β+2)=2k
α β + β+ 2α+2= 2k
α β + β+ α+α+2= 2k
k-2 + 2k +α+2= 2k
α=-k
put x=k into x²-2kx+(k-2)=0
k^2 + 2k^2 + k -2 = 0
3k^2 + k -2 =0
(3k-2)(k+1)=0
k= 2/3 or k = -1

Q1.若方程x-(2k-1)x+k=0 (k為實常數) 有不等實根α及β , 且α<β

(a)求k值的範圍

α+β=(2k+1) and αβ=k^2

b^2-4ac>0
(2k+1)^2-4(k^2)>0
4k+1>0
=>k>-1/4//

(b)用k表示αβ及α-β

αβ=(αβ)^2=k^4//

α-β=sqrt(α^2-2αβ+β^2)=sqrt((2k+1)^2-4k^2)=sqrt(4k+1)//


Q2.α、β為二次方程x-2kx+(k-2)=0的根.

(a)用k表示α+β及α ．β

α+β=2k and ααβ=k-2//

(α+1)(β+2)=2k
=>αβ+(α+β)+α=2k
=>k-2+(2k)+α+2=2k
=>α=-k

sub x==k ,we have
k^2+2(k)(k)+(k-2)=0
=>3k^2+k-2=0
=>(3k-2)(k+1)=0
=>k=2/3 or -1//


2009-01-01 23:44:45 補充：
(a)求k值的範圍

α+β=(2k+1) and αβ=k^2

b^2-4ac>0
(2k+1)^2-4(k^2)>0
-4k+1>0
=>k<1/4//

無所謂啦
1(a)
判別式>0
(2k-1)^2-4k^2>0
-4k+4>0
k<1
(b)
αβ
=(αβ)
=k^4
(α-β)^2
=(α+β)^2-4αβ
=(2k-1)^2-4k^2
=4-4k
(α-β)
=-sqrt(4-4k)
2
(a)
α+β=2k
α ．β=k-2
(b)
(α+1)(β+2)=2k
αβ+α+β+α+2=2k
k-2+2k+α+2=2k
α=-k
此時β=3k
-3k^2=k-2
3k^2+k-2=0
(3k-2)(k+1)=0
k=2/3或-1

This is clearly not a physics problem, why put it here ?